Recent content by threewingedfury

thats what I was thinking, I just didn't know because mult choice questions are so tricky! thanks!

The work in joules required to carry a 6.0 C charge from a 5.0 V equipotential surface to a 6.0V equipotential surface and back again to the 5.0V surface is: A) 0 B) 1.2 X 10^-5 C) 3.0 X 10^-5 D) 6.0 X 10^-5 E) 6.0X10^-6 I was thinkin the work is 0, but then again that seems too easy

m=r^2/R*lambda - .5 m=.017^2/(5*555*10^-9) - .5 m=104 m=(1.33).017^2/(5*555*10^-9) - .5 m=138 I've tried 103,104, and 105 for a and done the same for b but the method can't be right because in the book r=20mm, R=5m lambda=589nm and the answers are a. 34 and b. 46

Figure 35-46a shows a lens with radius of curvature R lying on a flat glass plate and illuminated from above by light with wavelength . Figure 35-46b, a photograph taken from above the lens, shows that circular interference fringes (called "Newton's rings") appear, associated with the variable...

so what would m be? do you add the two indecies together, or subtract them?

A sheet of glass having an index of refraction of 1.32 is to be coated with a film of material having an index of refraction of 1.46 such that green light with a wavelength of 525 nm (in air) is preferentially transmitted via constructive interference. (a) What is the minimum thickness of the...

So is it more like: (1/(4*π*ε_0)*(5.00 μC / (3.00 cm)^2) for particle 1 and give it an 'i' and -1/(4*π*ε_0)*(5.00 μC / (4.00 cm)^2) for particle 2 and give it a j? Since that's r/2?

pretty please? im a bit confused

so can anyone give me a better idea?

Two particles are fixed to an x-y coordinate system: particle 1 of charge -5.00 microCoulombs lies on the x-axis at x = + 6.00 cm and particle 2 of charge +5.00 microCoulombs lies on the y-axis at y = + 8.00 cm. Midway between the particles, what is their net electric field, in unit vector...

The intensity of sound wave A is 100 times that of sound wave B. Relative to wave B the sound level of wave A is: a. -2 dB b. +2 dB c. +10 dB d. +20 dB e. +100 db Using 10logbase10(I/Io) - 10log(100/1) - Would the answer be 20 dB?

in the center?

well I figured because it said the center of mass of the atmosphere that it didnt really include the earth

but I don't understand how the Earth and the atmosphere play roles in the question - I think of the Earth weighing so much, but it can't be the center of the Earth - I am all confused

so for the CG of the Earth's atmosphere - I don't really understand exactly how to figure this out. I figured it would be closer to the Earths surface, but is it halfway between the surface and the boundary of the atmosphere?